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8090 [49]
3 years ago
8

Which formula is comparing a sample of a gas under two different conditions of volume and temperature, if the pressure is consta

nt?
⚪︎ V1 V2

____ = ___

T1 T2


⚪︎ xy = k


⚪︎ V

__ = k

T


⚪︎ P1V1 = P2V2


⚪︎ PV = k
Chemistry
1 answer:
Gwar [14]3 years ago
3 0

Answer:

V1/T1 = V2/T2

Explanation:

The Combined gas law for two states, 1 and 2, is given by:

P1V1/T1 = P2V2/T2

where P, V, and T are Pressure, Volume, and Temperature (always in degrees Kelvin).

If pressure is constant, the P1 and P2 cancel out:

V1/T1 = V2/T2

I can't tell from the formatting of the answers if A is this relationship.  Please compare.

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A. 207 kJ<br> B. 4730 kJ<br> O C. 9460 kJ<br> O D. 414 kJ
slavikrds [6]

Answer:

C = 9460 Kj

Explanation:

Given data:

Mass of copper = 2kg

Latent heat of vaporization = 4730 Kj/Kg

Energy required to vaporize 2kg copper = ?

Solution:

Equation

Q= mLvap

by putting values,

Q= 2kg ×  4730 Kj/Kg

Q = 9460 Kj

3 0
3 years ago
The lowest possible concentration of free hydrogen ions is represented by a ph value of
spayn [35]
PH is the logarithmic measure of the concentration of hydrogen ions in solution. In an aqueous system, the lowest possible concentration of H+ ions (least acidic) is 1x 10^-14. The -log(1x10^-14) = pH of 14
6 0
3 years ago
How many grams of calcium phosphate (Ca3(PO4)2) re theoretically produced if we start with 3.40 moles of Ca(NO3)2 and 2.40moles
sattari [20]

1) Balance the chemical equation.

3Ca(NO_3)_2+2Li_3PO_4\rightarrow6LiNO_3+Ca_3(PO_4)_2

2) List the known and unknown quantities.

Reactant 1: Ca(NO3)2.

Amount of substance: 3.40 mol.

Reactant 2: Li3PO4.

Amount of substance: 2.40 mol.

Product: Ca3(PO4)2

Mass: unknown.

3) Which is the limiting reactant?

<em>3.1-How many moles of Li3PO4 do we need to use all of the Ca(NO3)2?</em>

The molar ratio between Li3PO4 and Ca(NO3)2 is 2 mol Li3PO4: 3 mol Ca(NO3)2.

mol\text{ }Li_3PO_4=3.40\text{ }mol\text{ }Ca(NO_3)_2*\frac{2\text{ }mol\text{ }Li_3PO_4}{3\text{ }mol\text{ }Ca(NO_3)_2}=2.2667\text{ }mol\text{ }Li_3PO_4

<em>We need 2.2667 mol Li3PO4 and we have 2.40 mol Li3PO4. We have enough Li3PO4. </em>This is the excess reactant.

<em>3.2-How many moles of Ca(NO3)2 do we need to use all of the Li3PO4?</em>

The molar ratio between Li3PO4 and Ca(NO3)2 is 2 mol Li3PO4: 3 mol Ca(NO3)2.

mol\text{ }Ca(NO_3)_2=2.40\text{ }mol\text{ }Li_3PO_4*\frac{3\text{ }mol\text{ }Ca(NO_3)_2}{2\text{ }mol\text{ }Li_3PO_4}=3.60\text{ }mol\text{ }Ca(NO_3)_2

<em>We need 3.60 mol Ca(NO3)2 and we have 3.40 mol Ca(NO3)2. We do not have enough Ca(NO3)2. </em>This is the limiting reactant.

4) Moles of Ca3(PO4)2 produced from the limiting reactant.

We have 3.40 mol Ca(NO3)2 of the limiting reactant.

The molar ratio between Ca(NO3)2 and Ca3(PO4)2 is 3 mol Ca(NO3)2: 1 mol Ca3(PO4)2.

mol\text{ }Ca_3(PO_4)_2=3.40\text{ }mol\text{ }Ca(NO_3)_2*\frac{1\text{ }mol\text{ }Ca_3(PO_4)_2}{3\text{ }mol\text{ }Ca(NO_3)_2}=1.1313\text{ }mol\text{ }Ca_3(PO_4)_2

5) Mass of Ca3(PO4)2 produced.

The molar mass of Ca3(PO4)2 is 310.1767 g/mol.

g\text{ }Ca_3(PO_4)_2=1.1333\text{ }mol\text{ }Ca_3(PO_4)_2*\frac{310.1767\text{ }g\text{ }Ca_3(PO_4)_2}{1\text{ }mol\text{ }Ca_3(PO_4)_2}g\text{ }Ca_3(PO_4)_2=351.526\text{ }g\text{ }Ca_3(PO_4)_2

<em>The mass of Ca3(PO4)2 produced is</em> 351 g Ca3(PO4)2.

Option D.

.

8 0
1 year ago
How is a coefficient used to balance an equation
ludmilkaskok [199]

Let's start to understand this question by a simple combustion reaction involving oxidation of Ethane in the presence of Oxygen. When Ethane is burned in the presence of Oxygen it produces Carbon Dioxide and Water respectively. Therefore, the equation is as,

                                C₂H₆  +  O₂    →    CO₂  +  H₂O

Above reaction shows the reaction and the equation is unbalanced. Balancing chemical equation is important because according to law of conservation of mass, mass can neither be created nor destroyed. Hence, we should balance the number of elements on both side.

                                       LHS                      RHS

Carbon Atoms                  2                            1

Hydrogen Atoms              6                           2

Oxygen Atoms                  2                           3

It means this equation is not obeying the law. Now, how to balance? One way is as follow,

                                C₂H₆  +  O₃    →    C₂O₂  +  H₆O

                                       LHS                      RHS

Carbon Atoms                  2                            2

Hydrogen Atoms              6                           6

Oxygen Atoms                  3                           3

We have balanced the equation by changing the subscripts. But, we have messed up the chemical composition of compounds and molecules like Oxygen is converted into Ozone.

Therefore, we will change the coefficients (moles) to balance the equation as,

                                C₂H₆  +  7/2 O₂    →    2 CO₂  +  3 H₂O

                                       LHS                      RHS

Carbon Atoms                  2                            2

Hydrogen Atoms              6                           6

Oxygen Atoms                  7                           7

Now, by changing the coefficients we have balanced the equation without disturbing the chemical composition of compounds and molecules.

3 0
3 years ago
Kernel structure for manganese
Amanda [17]

Answer:

The nucleus consists of 25 protons (red) and 30 neutrons (blue). 25 electrons (green) bind to the nucleus, successively occupying available electron shells (rings). Manganese is a transition metal in group 7, period 4, and the d-block of the periodic table. It has a melting point of 1246 degrees Celsius.

3 0
3 years ago
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